Question

f(x)在[0,1]上连续,且满足f(x)=e^x+∫[1,0]2f(t)dt,求f(x)

Answer 1

ƒ(x) = e^x + ∫(0→1) 2ƒ(t) dt

令常数项∫(0→1) ƒ(t) dt = A,于是

ƒ(x) = e^x + 2A,两边取定积分

∫(0→1) ƒ(x) dx = ∫(0→1) (e^x + 2A) dx

A = [e^x + 2Ax] |(0→1) = (e + 2A) - (1 + 0) = e + 2A - 1

- A = e - 1

A = 1 - e

所以ƒ(x) = e^x + 2(1 - e)

ƒ(x) = e^x - 2e + 2